Balance the following chemical equation: $ $ $\text{Al} +$ $\text{O}_2 \rightarrow$ $\text{Al}_2\text{O}_3$
Solution: There are $2 \text{ O}$ on the left and $3$ on the right. The lowest common denominator is $6$ , so multiply $\text{O}_2$ by ${3}$ and $\text{Al}_2\text{O}_3$ by ${2}$ $ \text{Al} + {3}\text{O}_2 \rightarrow {2}\text{Al}_2\text{O}_3 $ That gives us $4 \text{ Al}$ on the right and only $1$ on the left, so multiply $\text{Al}$ by ${4}$ $ {4}\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 $ The balanced equation is: $ 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 $